Correct me if I'm wrong but isn't Average Force defined simply as delta p/delta t. Since the stone comes to rest, assuming all momentum is transfered to the nail, the momentum of the stone just before it hits the nail is delta p. For delta t we can use the average v which is (10-0)/2 = 5m/s. Then
delta t = y/(average v) =0.025/5.
Plugging this in:
Average F = (5kg)(10m/s)(5m/s)/(25x10^-3m)=10^4N.

I think this question is somehow vague. Actually the time average and position average are not generally the same. But this question says just average force and it is not clear which average it means. However, the given information of the question is adequate only for calculation of position average which is . I don\'t see any way to calculate time average of the force which is by using the given information of this question.

Your zero potential energy ought to be the surface of the wood. If you are concerned about the change in gravitational potential energy of the nail since it travels downward, understand that it is negligible. You would need a ridiculously huge nail for it to affect the answer. Also note that the question asks for what the average force "most nearly" is.

But this question asked for the average force (Fa)!
Fa = (Force final + Force initial)/2
Force initial is the Force on initial impact (20,000N) and Force final is the force exerted immediately after the nail stops moving (0N). so....

Correct me if I'm wrong but isn't Average Force defined simply as delta p/delta t. Since the stone comes to rest, assuming all momentum is transfered to the nail, the momentum of the stone just before it hits the nail is delta p. For delta t we can use the average v which is (10-0)/2 = 5m/s. Then
delta t = y/(average v) =0.025/5.
Plugging this in:
Average F = (5kg)(10m/s)(5m/s)/(25x10^-3m)=10^4N.

Imperate2008-09-11 12:23:00

This is the way I did it.rnIf the block has momentum Then this momentum must be reduced to zero over the course of the nail being driven in. The time it the nail travels can be found from Which implies rnrnSo it can be done this way, but I wish I thought of using energy at the time.

Setareh2011-10-24 00:30:23

I doubt whether you are allowed to use this equation : t=y/v or not, since we don't deal with constant speed motion, actually we have acceleration, so this equation is not valid in this case.

kronotsky2018-10-23 02:46:41

This solution is essentially the same as the one given, because in both cases the assumption is made that the nail experiences a constant force. This is the correct approach for the time average of the force, but we still need to know the time dependence. Similarly, conservation of energy is the right approach to find the distance average of the force.

I think it would be better phrased as a Work Problem. It's not really energy conservation, because energy is clearly not conserved.
Anyway, not that it makes any difference to the solution really, but you forgot to include the potential energy term . It only contributes 50 J or so...so who cares right?

I have to disagree with everyone. Kinematics give the average net force as 100000/9.81 N 10190 N (downward), but the question asks for the average net force exerted BY the stone on the nail. (The binomial theorem will be useful to make this approximation). The average frictional force acting on the nail is = 10000N (upward) from the expression for the mechanical energy lost due to a constant dissipative force. The magnitude of the force exerted on the nail BY the stone is therefore approximately 190 N. As an order of magnitude answer, then, the best anwer is B.

Go maroons!

kicksp2007-10-29 13:05:19

Sorry, I read the problem wrong. the average force exerted ON the stone BY the nail is 10000 N:

By Newton's Third Law, this is the magnitude of the average force exerted ON the nail BY the stone. Still, the analysis by energy conservation is erroneous.

I have to disagree with everyone. Kinematics give the average net force as 100000/9.81 N 10190 N (downward), but the question asks for the average net force exerted BY the stone on the nail. (The binomial theorem will be useful to make this approximation). The average frictional force acting on the nail is = 10000N (upward) from the expression for the mechanical energy lost due to a constant dissipative force. The magnitude of the force exerted on the nail BY the stone is therefore approximately 190 N. As an order of magnitude answer, then, the best anwer is B.

First of all, I do not believe we can apply conservation of energy to this problem. It's possible that its just really late and I am really tired, but it seems to me that energy isn't necessarily conserved here since friction is slowing down the nail and dissipating energy in heat and sound. Additionally, the problem does not say that the collision between the stone and the nail is elastic so I see no reason to attack this problem from a conservation of energy foundation.

I do believe that conservation of momentum, though, is the way to go. I used two equations in my solution, x=v0t-1/2at^2 and J = Ft = delta p = m delta v.

Given that x = .025, delta v = 10 = v0 and F=ma, I solved for t in the impulse equation and plugged into the kinematic eqn to solve for F.

hungrychemist2007-08-26 19:18:40

This question cannot be solved exactly. Your kinematic approach is only approximation since one cannot assume the constant acceleration of the pin. Like Yosun said, energy conservation approach is the way to go although energy is not really conserved here you can see from the choices, within a reasonable loss of energy due to (friction and sound and so on) you can conclude the ans to be D.

TigerTed82007-09-17 19:36:41

Karsten, I did the problem the same way as you when I first approached it. However, I'm now a convert to the energy method. The friction of the nail, clearly non-trivial, is not a part of the equation here. The stone has energy, and then it loses that energy. This occurs primarily through some force. So = . The only way that the stone could lose energy that wasn't related to a force would be IR radiation, or the sound or heat from the stone on nail collision. The nail-wood friction occurs AFTER the transfer of energy from stone to nail by way of . Hope this helps.

kronotsky2018-10-23 02:49:28

Who cares if energy is conserved? The energy lost by the nail is due to the forces on it.